A small bob mass `'m'` is suspended by a massless string from a cart of the same mass `'m'` as shown in the figure. The friction between the cart and horizontal ground is negligible. The bob is given a velocity `V_(0)` in horizontal direction as shown. The maximum height attained by the bob is , `(` initially whole system `(` bob `+` string `+` cart `)` was at rest `)`.

Options

A. `(2V_(0)^(2))/g`
B. `(V_(0)^(2))/g`
C. `(V_(0)^(2))/(4g)`
D. `(V_(0)^(2))/(2g)`

Answer

C

Explanation

A small bob of
By linear momentum conservation in horizontal direction = for (bot +string+cart)
`v=(v_(0))/2`
By mechanical energy conservation for
(bob+string +cart +earth)
`1/2mV_(a)^(2)+0+0=1/2(2m)v^(2)+mgh+0`
`1/2mV_(0)^(2)-1/2(2m)(V_(0)^(2))/4=mgh`
Solving it `h=(V_(0)^(2))/(4g)`